Posted by **Anonymous** on Friday, March 7, 2014 at 9:26pm.

Two forces,

1 = (3.85 − 2.85) N

and

2 = (2.95 − 3.65) N,

act on a particle of mass 2.10 kg that is initially at rest at coordinates

(−2.30 m, −3.60 m).

(a) What are the components of the particle's velocity at t = 11.8 s?

= m/s

(b) In what direction is the particle moving at t = 11.8 s?

° counterclockwise from the +x-axis

(c) What displacement does the particle undergo during the first 11.8 s?

Δ = m

(d) What are the coordinates of the particle at t = 11.8 s?

x = m

y = m

- physics -
**Henry**, Monday, March 10, 2014 at 11:06pm
F1 = (x,y) = (3.85N,-2.85N)

F2 = (x,y) = (2.95N,-3.65N).

a. Fr=X+Yi=(3.85+2.95) + (-2.85i-3.65i)=

6.8 - 6.5i = 9.41N[316.3o] = Resultant

force.

2.10kg particle

Location: (x,y) = (-2.30m,-3.60m)

a(x) = Fx/mass = 6.8/2.1 = 3.24 m/s^2

Vx=Vox + a*t = 0 + 3.24*11.8= 38.21 m/s.

a(y) = Fy/mass = -6.5/2.1 = -3.10 m/s^2

Vy=Voy + a*t = 0 - 3.1*11.8= 36.52 m/s.

b. Tan A = Vy/Vx = 36.52/38.21=0.95587

A = 43.71o

c. d(x) = (Vx^2-Vox^2)/2a =

(38.21^2-0)/6.48 = 225.3 m.

Dx = 225.3 - (-2.30) = 227.6 m. = Hor.

comp. of displacement.

d(y) = (Vy^2-Voy^2)/2a =

(36.52^2-0)/-6.2 = -215.1 m.

Dy = -215.1 - (-3.60) = -211.5 m. = Ver.

comp. of displacement.

Tan Ar = Dy/Dx = -211.5/227.6 = -0.92933

Ar = -42.9o

Displacement=Dx/cosA=227.6/cos42.9=310.7

m.

X = 225.3 m

Y = -215.1 m.

d. (d(x),d(y)) = (225.3m,-215.1m)

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